(i) Using the equation of motion \(s = ut + \frac{1}{2}at^2\), where \(u = 0\), \(s = 4.5\), and \(t = 5\), we have:

\(4.5 = 0 + \frac{1}{2} \times a \times 5^2\)

Solving for \(a\), we get \(a = 0.36 \text{ m/s}^2\).

Resolving horizontally, the net force is \(6 \times \frac{24}{25} - F = 3 \times 0.36\).

Solving for \(F\), we find \(F = 4.68 \text{ N}\).

(ii) The normal reaction \(R\) is given by:

\(R = 3g - 6 \sin 16.3 = 3g - 6 \times \frac{7}{25} = 28.32 \text{ N}\)

Using \(F = \mu R\), we have:

\(4.68 = \mu \times 28.32\)

Solving for \(\mu\), we get \(\mu = 0.165\).

(iii) The velocity at \(t = 5\) is \(v = 5 \times 0.36 = 1.8 \text{ m/s}\).

Using \(3a = -0.165 \times 3g\), we solve for \(t\) when \(v = 0\):

\(0 = 1.8 - 0.165g t\)

Solving for \(t\), we find \(t = 1.09 \text{ s}\).

The total time is \(5 + 1.09 = 6.09 \text{ s}\).